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Joined: 10 Apr 2015
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√(2x + 2) −√(x − 3) = 2
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07 Feb 2022, 07:17
07 Feb 2022, 07:17
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\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x\) |
\(7\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
√(2x + 2) −√(x − 3) = 2
[#permalink]
07 Feb 2022, 17:06
07 Feb 2022, 17:06
GreenlightTestPrep wrote:
\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
Quantity A |
Quantity B |
\(x\) |
\(7\) |
The equation implies that 2x+2 and x-3 are both PERFECT SQUARES, since the difference between their square roots is an INTEGER.
Further, 2x+2 and x-3 must be two perfect squares that are relatively close to each other, since the difference between their square roots is quite small (2).
Only the value given for Quantity B -- 7 -- will render for 2x+2 and x-3 two perfect squares that are relatively close to each other:
2x+2 = 2*7 + 2 = 16
x-3 = 7-3 = 4
Implication:
x=7 is the only possible solution for the equation, with the result that Quantities A and B are equal.
Show: ::
C
√(2x + 2) −√(x − 3) = 2
[#permalink]
07 Feb 2022, 17:24
07 Feb 2022, 17:24
A useful identity:
\((a+b)^2 = a^2 + 2ab + b^2\)
Algebraic solution:
Adding \(\sqrt{x − 3}\) to both sides, we get:
\(\sqrt{2x + 2}=\sqrt{x − 3} + 2\)
Squaring both sides, we get:
\(2x+2 = (x-3) + (2*\sqrt{x−3}*2) + 2^2\)
\(2x+2 = x+1+4\sqrt{x − 3}\)
Subtracting x+1 from both sides so that the root is more isolated, we get:
\(x+1=4\sqrt{x − 3}\)
Squaring both sides, we get:
\(x^2 + 2*x*1 + 1^2 = 16(x-3)\)
\(x^2 + 2x + 1 = 16x-48\)
\(x^2-14x+49 = 0\)
\((x-7)^2 = 0\)
\(x=7\)
Thus, Quantities A and B are equal.
\((a+b)^2 = a^2 + 2ab + b^2\)
GreenlightTestPrep wrote:
\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
Quantity A |
Quantity B |
\(x\) |
\(7\) |
Algebraic solution:
Adding \(\sqrt{x − 3}\) to both sides, we get:
\(\sqrt{2x + 2}=\sqrt{x − 3} + 2\)
Squaring both sides, we get:
\(2x+2 = (x-3) + (2*\sqrt{x−3}*2) + 2^2\)
\(2x+2 = x+1+4\sqrt{x − 3}\)
Subtracting x+1 from both sides so that the root is more isolated, we get:
\(x+1=4\sqrt{x − 3}\)
Squaring both sides, we get:
\(x^2 + 2*x*1 + 1^2 = 16(x-3)\)
\(x^2 + 2x + 1 = 16x-48\)
\(x^2-14x+49 = 0\)
\((x-7)^2 = 0\)
\(x=7\)
Thus, Quantities A and B are equal.
Show: ::
C
